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anonymous

  • 5 years ago

If y = 5/2(e^(x/5) + e^(-x/5)), prove that y'' = y/25.

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  1. anonymous
    • 5 years ago
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    I expanded the bracket: y = 5/2(e^(x/5)) + 5/2(e^(-x/5))

  2. anonymous
    • 5 years ago
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    Then I took the first derivative:

  3. anonymous
    • 5 years ago
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    y ' = 5x/10(e^(x/5)) - 5x/10(e^(-x/5))

  4. anonymous
    • 5 years ago
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    Now how should I go about with getting the second derivative? Should I factor out the 5x/10 and then use product rule?

  5. anonymous
    • 5 years ago
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    \[y'=5/2[e ^{x/5}(1/5)+e^{-x/5}(-1/5)\]

  6. anonymous
    • 5 years ago
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    \[y'=(5/2)(1/5)[e ^{x/5}-e ^{-x/5}]\]

  7. anonymous
    • 5 years ago
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    Does it make a difference if we were to expand first with 5/2, and then take the first derivaitve?

  8. anonymous
    • 5 years ago
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    not at all

  9. anonymous
    • 5 years ago
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    Because that is what I did and then I took the dy/dx of each of the two terms.

  10. anonymous
    • 5 years ago
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    but y did u multiply with x/5?

  11. anonymous
    • 5 years ago
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    instead it is 1/5 the coefficient of x/5

  12. anonymous
    • 5 years ago
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    No sorry I did the following: y = 5/2(e^{x/5}) + 5/2(e^{-x/5}) y '= 5/2(1/5)*(e^{x/5}) + 5/2(-1/5)(e^{-x/5})

  13. anonymous
    • 5 years ago
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    correct

  14. anonymous
    • 5 years ago
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    So then: y = 1/2(e^{x/5} + e^{-x/5})

  15. anonymous
    • 5 years ago
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    y' u mean?

  16. anonymous
    • 5 years ago
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    yes sorry, y'

  17. anonymous
    • 5 years ago
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    y' = 1/2(e^{x/5} + e^{-x/5})

  18. anonymous
    • 5 years ago
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    n there must b negative between to terms

  19. anonymous
    • 5 years ago
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    coz 1/5 is the common factor

  20. anonymous
    • 5 years ago
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    y' = 1/2(e^{x/5} - e^{-x/5}), right?

  21. anonymous
    • 5 years ago
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    yes:)

  22. anonymous
    • 5 years ago
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    So then for y'' I get..

  23. anonymous
    • 5 years ago
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    \[y''=1/2[e ^{^{x/5}}(1/5)-e ^{-x/5}(-1/5)]\]

  24. anonymous
    • 5 years ago
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    Yes I got that, then I expanded, and I got: y'' = 1/10(e^{x/5} + e^{-x/5})

  25. anonymous
    • 5 years ago
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    \[y''=1/10[e ^{x/5}+e ^{-x/5}]\]

  26. anonymous
    • 5 years ago
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    again no need to expand

  27. anonymous
    • 5 years ago
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    \[e ^{x/5}+ e ^{-x/5}=2y/5\]

  28. anonymous
    • 5 years ago
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    y''=y/25

  29. anonymous
    • 5 years ago
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    How did you get e^x/5 + e^-x/5 = 2y/5?

  30. anonymous
    • 5 years ago
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    OHHH

  31. anonymous
    • 5 years ago
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    I think I know

  32. anonymous
    • 5 years ago
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    got it?

  33. anonymous
    • 5 years ago
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    I got it thank you so much. So when you are taking derivative of a function that has a constant which is multipled by a bracket that can be differentiated, you just take the derivative of the terms inside the bracket and multiply by the constant in front of it?

  34. anonymous
    • 5 years ago
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    right:)

  35. anonymous
    • 5 years ago
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    Ok awesome. Thank you for your help.

  36. anonymous
    • 5 years ago
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    my pleasure

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